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The sum of the positive integers between 99 and 999 is 299500

Since a = 100, and d = 2 note that the last term will be 998 using nth tem of an Ap

Un= a+(n-1)d

998= 100+ (n-1)2

Then 998- 100= (n- 1)2

= 898= (n-1)2

449= n-1

n= 450

Sum of the terms = n/2(2a+(n-1)d)

450/2(2(100) + 449*2)

Ans= 247050

=

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